GCSE nth term of quadratic sequences Practice Questions

Free GCSE nth term of quadratic sequences practice questions with full step-by-step worked solutions. Covers second differences, quadratic sequences, coefficient of n squared, fractional coefficient. Practise exam-style problems and check your method.

second differencesquadratic sequencescoefficient of n squaredfractional coefficientsubstitutionquadratic nth term
GCSE Higher70 questionsStep-by-step solutions
Question 1
1 markeasy
Here are the first five terms of a quadratic sequence. 3, 7, 13, 21, 31, …3,\ 7,\ 13,\ 21,\ 31,\ \ldots Work out the second difference of the sequence.
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Worked solution

  1. Work out the first differences

    4, 6, 8, 104,\ 6,\ 8,\ 10

    Take each term away from the term after it.

  2. Work out the second differences

    2, 2, 22,\ 2,\ 2

    Now difference the first differences.

  3. State the second difference

    22

    The second difference is constant, which confirms the sequence is quadratic.

Answer
22
Question 2
2 markseasy
A geometric sequence begins 5, 5, 55, 25, …\sqrt{5},\ 5,\ 5 \sqrt{5},\ 25,\ \ldots Work out the next term. Give your answer in surd form.
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Worked solution

  1. Find the common ratio

    r=55=5r = \frac{5}{\sqrt{5}} = \sqrt{5}

    Rationalising, 55=555=5\frac{5}{\sqrt{5}} = \frac{5\sqrt{5}}{5} = \sqrt{5}.

  2. Check the ratio further along

    2555=55=5\frac{25}{5\sqrt{5}} = \frac{5}{\sqrt{5}} = \sqrt{5}

    The ratio is constant, so the sequence is geometric with r=5r = \sqrt{5}.

  3. Multiply the last term by the ratio

    25×5=25525 \times \sqrt{5} = 25\sqrt{5}

    The 5th term is 25525\sqrt{5}.

Answer
25525\sqrt{5}
Question 3
2 marksintermediate
A geometric sequence has first term 44 and common ratio 12\frac{1}{2}. Work out the 66th term. Give your answer as a fraction.
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Worked solution

  1. Write the rule for the kth term

    uk=4×(12)k−1u_k = 4 \times \left(\frac{1}{2}\right)^{k-1}

    First term a=4a = 4, ratio r=12r = \frac{1}{2}.

  2. Substitute k equals 6

    u6=4×(12)5u_6 = 4 \times \left(\frac{1}{2}\right)^{5}

    Five halvings take you from the 1st term to the 6th.

  3. Work out the power

    (12)5=132\left(\frac{1}{2}\right)^{5} = \frac{1}{32}

    25=322^5 = 32.

  4. Multiply

    4×132=4324 \times \frac{1}{32} = \frac{4}{32}

    Multiply the numerator by 44.

  5. Simplify

    432=18\frac{4}{32} = \frac{1}{8}

    Divide top and bottom by 44.

  6. State the term

    18\frac{1}{8}

    Listing the terms 4, 2, 1, 12, 14, 184,\ 2,\ 1,\ \frac{1}{2},\ \frac{1}{4},\ \frac{1}{8} confirms it.

Answer
18\frac{1}{8}
Question 4
3 markshard
Here are the first five terms of a quadratic sequence. 7, 18, 35, 58, 87, …7,\ 18,\ 35,\ 58,\ 87,\ \ldots Find an expression, in terms of nn, for the nnth term of the sequence.
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Worked solution

  1. Write out the terms

    7, 18, 35, 58, 877,\ 18,\ 35,\ 58,\ 87

    These are the first 5 terms of the sequence.

  2. Work out the first differences

    11, 17, 23, 2911,\ 17,\ 23,\ 29

    Subtract each term from the one after it. The first differences are not constant, so the sequence is not linear.

  3. Work out the second differences

    6, 6, 66,\ 6,\ 6

    Now take the differences of the first differences.

  4. Confirm the sequence is quadratic

    second difference=6\text{second difference} = 6

    The second difference is constant and not zero, so the sequence is quadratic: its nnth term has the form an2+bn+can^2 + bn + c.

  5. Halve the second difference to get the coefficient of n squared

    a=62=3a = \frac{6}{2} = 3

    The second difference of an2+bn+can^2+bn+c is always 2a2a, so aa is HALF the second difference. Forgetting to halve is the classic error here.

  6. Write out a times n squared for each position

    3, 12, 27, 48, 753,\ 12,\ 27,\ 48,\ 75

    These are the values of 3n23 n^{2} for n=1,2,3,…n = 1, 2, 3, \ldots

  7. Subtract a times n squared from each term

    4, 6, 8, 10, 124,\ 6,\ 8,\ 10,\ 12

    What is left is the linear part bn+cbn + c of the rule.

  8. Find the linear part

    bn+c:b=2, c=2bn + c: \quad b = 2,\ c = 2

    The leftover sequence goes up by 22 each time, so b=2b = 2. Its first value is 44, and 2+c=42 + c = 4 gives c=2c = 2.

  9. Write down the nth term

    3n2+2n+23 n^{2} + 2 n + 2

    Putting the quadratic part and the linear part together gives the rule for the nnth term.

  10. Check the rule against a given term

    n=5:3×52+2×5+2=87n = 5: \quad 3 \times 5^2 + 2 \times 5 + 2 = 87

    Substituting n=5n = 5 regenerates the given term 8787, so the rule is right.

Answer
3n2+2n+23 n^{2} + 2 n + 2
Question 5
5 markschallenging
A quadratic sequence has nnth term an2+bn+can^2 + bn + c. Its first three terms are T1T_1, T2T_2 and T3T_3. Which one of these expressions always gives the value of aa?
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Worked solution

  1. Write the first three terms in terms of a, b and c

    T1=a+b+c,T2=4a+2b+c,T3=9a+3b+cT_1 = a + b + c, \quad T_2 = 4a + 2b + c, \quad T_3 = 9a + 3b + c

    Substitute n=1,2,3n = 1, 2, 3 into an2+bn+can^2 + bn + c.

  2. Work out the first differences

    T2−T1=3a+b,T3−T2=5a+bT_2 - T_1 = 3a + b, \qquad T_3 - T_2 = 5a + b

    Subtract consecutive terms.

  3. Work out the second difference

    (T3−T2)−(T2−T1)=(5a+b)−(3a+b)=2a(T_3 - T_2) - (T_2 - T_1) = (5a + b) - (3a + b) = 2a

    The bb terms cancel.

  4. Expand the second difference in terms of the T values

    T3−2T2+T1=2aT_3 - 2T_2 + T_1 = 2a

    Rearranging (T3−T2)−(T2−T1)(T_3 - T_2) - (T_2 - T_1) gives T1−2T2+T3T_1 - 2T_2 + T_3.

  5. Solve for a

    a=T1−2T2+T32a = \frac{T_1 - 2T_2 + T_3}{2}

    Divide the second difference by 22 -- the halving step.

  6. Check the coefficients directly

    T1−2T2+T3=(1−8+9)a+(1−4+3)b+(1−2+1)c=2aT_1 - 2T_2 + T_3 = (1 - 8 + 9)a + (1 - 4 + 3)b + (1 - 2 + 1)c = 2a

    The bb and cc coefficients are both zero, so only aa survives.

  7. Test with a known example

    un=n2:T1=1,T2=4,T3=9u_n = n^2: \quad T_1 = 1, T_2 = 4, T_3 = 9

    Here a=1a = 1, so the formula should give 11.

  8. Apply the formula to the example

    1−8+92=22=1✓\frac{1 - 8 + 9}{2} = \frac{2}{2} = 1 \quad \checkmark

    The formula returns a=1a = 1, as it should.

  9. Show the unhalved version fails

    T1−2T2+T3=2≠1T_1 - 2T_2 + T_3 = 2 \ne 1

    Forgetting to halve gives 22, which is 2a2a, not aa. This is the classic error.

  10. Show a difference-based distractor fails

    T3−T12=9−12=4≠1\frac{T_3 - T_1}{2} = \frac{9 - 1}{2} = 4 \ne 1

    This is not aa either.

  11. Show the first-difference distractor fails

    T2−T1=4−1=3≠1T_2 - T_1 = 4 - 1 = 3 \ne 1

    The first difference is 3a+b3a + b, not aa.

  12. Show the halved first difference fails

    T2−T12=32≠1\frac{T_2 - T_1}{2} = \frac{3}{2} \ne 1

    Halving the first difference is not the same as halving the second difference.

  13. Test the correct formula on a second example

    un=−2n2+n:T1=−1,T2=−6,T3=−15u_n = -2n^2 + n: \quad T_1 = -1, T_2 = -6, T_3 = -15

    Here a=−2a = -2.

  14. Apply the formula again

    −1−2(−6)+(−15)2=−1+12−152=−42=−2✓\frac{-1 - 2(-6) + (-15)}{2} = \frac{-1 + 12 - 15}{2} = \frac{-4}{2} = -2 \quad \checkmark

    It works for a negative aa too.

  15. State the correct expression

    a=T1−2T2+T32a = \frac{T_1 - 2T_2 + T_3}{2}

    This is half the second difference, written directly in terms of the first three terms.

Answer
a=T1−2T2+T32a = \frac{T_1 - 2T_2 + T_3}{2}

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